`21=3.7`

`44=2^{2}.11`

`14=2.7`

`33=3.11`

  `=>BCN N(21,44,14,33)=3.7.2^{2}.11=924`

924 là BCNN

[x-2].[x mũ 2 - 16]=0

[x-2]-[x mũ 2 - 16] = 0

TH1:   x-2=0

           x=0+2

           x=2[thỏa mãn]

TH2:    x mũ 2 - 16=0

            x mũ 2=0+16

            x mũ 2=   16

            x mũ 2=4 mũ 2          [nghĩa là 16= 4 mũ 2]

            x=4

          Vậy....

phần b làm giống thế nha

-351 - 51( x - 2 ) = -759

51( x - 2 ) = ( -351 ) - ( -759 ) = 408

x - 2 = 408 : 51 = 8

x = 8 + 2 = 10

-321-51x[x-2]=-759

51x[x-2]=-321-[-759]

51x[x-2]=408

x-2=408:51

x-2=8

x=8+2

x=10

Vậy x=10

=\(\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}\)

=\(\dfrac{5}{11}\times1+\dfrac{6}{11}\)

=\(\dfrac{11}{11}\)=1

\(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)

B = 2-4-6+8 + 10-12-14+16 + ... + 90-92-94+96 + 98 - 100

B =  (2-4-6+8) + (10-12-14+16) + ... + (90-92-94+96) + (98 - 100)

B = 0 + 0 + ... + 0 + (-2)

B = -2

A = | x| + 2003 

|x| ≥ 0 ⇒ |x| + 2003 ≥ 2003  

A(min) = 2003 khi x = 0

1. 

$-(a+b-c)-(a-b-c)+(a-b+c)$

$=-a-b+c-a+b+c+a-b+c$

$=(-a-a+a)+(-b+b-b)+(c+c+c)$

$=-a-b+3c$

2.

$-(-a-b-c)-(a-b+c)-(a+b+c)$

$=a+b+c-a+b-c-a-b-c$

$=(a-a-a)+(b+b-b)+(c-c-c)$

$=-a+b-c$

3.

$-(-a+b+c)-(-a-b-c)-(2a-b+c)$

$=a-b-c+a+b+c-2a+b-c$

$=(a+a-2a)+(-b+b+b)+(-c+c-c)$

$=b-c$

4.

$(2a+3b-c)-(2a-b-2c)-(c-b+a)$

$=2a+3b-c-2a+b+2c-c+b-a$

$=(2a-2a-a)+(3b+b+b)+(-c+2c-c)$

$=-a+5b$

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100

2A = 1 + 1/2 + 1/2^2 + ... + 1/2^99
2A - A = (1 + 1/2 + 1/2^2 + ... + 1/2^99) - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100)

A = 1 - 1/2^100

B = 1 + 1/3 + 1/3^3 + ... + 1/3^2022

3B = 3 + 1 + 1/3 + ... + 1/3^2021

3B - B = (3 + 1 + 1/3 + ... + 1/3^2021) - (1 + 1/3 + 1/3^3 + ... + 1/3^2022)

2B = 3 - 1/3^2022

B = \(\dfrac{\text{3 - 1/3^2022}}{\text{2}}\)

   A        =      \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) +...............+ \(\dfrac{1}{2^{100}}\)

2.A       = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) +\(\dfrac{1}{2^3}\).........+\(\dfrac{1}{2^{99}}\)

2A -A   =  1 - \(\dfrac{1}{2^{100}}\)

       A  =   1 - \(\dfrac{1}{2^{100}}\)

B          =    1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^3}\) + ....+ \(\dfrac{1}{3^{2022}}\) 

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